Bài 3:

a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)

\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)

Vì \(3\ne0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)

b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)

c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)

\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)

Chúc bạn học tốt!

b) Ta có: \(x^3+4x+5=0\)

\(\Leftrightarrow x^3-x+5x+5=0\)

\(\Leftrightarrow x\left(x^2-1\right)+5\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+5\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+5\right)=0\)

mà \(x^2-x+5>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

a)x²-(x+3)(3x+1)=9

⇔(x-3)(x+3)-(x+3)(3x+1)=0

⇔x+3=0 hoặc 3x+1=0 

1.x+3=0 ⇔x=-3

2.3x+1=0⇔x=-1/3

phương trình có 2 nghiệm x=-3 và x=-1/3

Áp dụng công thức: \(A\left(x\right).B\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}A\left(x\right)=0\\B\left(x\right)=0\end{matrix}\right.\)

a) \(PT\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)

b) \(PT\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)

Vậy: \(S=\left\{3;20\right\}\)

c) Vì \(x^2+1\ge1>0\forall x\)

\(\Rightarrow4x+2=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

d) \(PT\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)

a: =>3x-2=0 hoặc 4x+5=0

=>x=2/3 hoặc x=-5/4

b: =>(x-3)(x+20)=0

=>x=3 hoặc x=-20

c: =>4x+2=0

hay x=-1/2

d: =>2x+7=0 hoặc x-5=0 hoặc 5x+1=0

=>x=-7/2 hoặc x=5 hoặc x=-1/5

a: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)

f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)

g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)

b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)

c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)